A bag contains 21 toys numbered 1 to 21. A toy is drawn and then another toy is drawn without replacement. Find the probability that both toys will show even numbers.

Answer: Option B

The probability that first toy shows the even number $$=\frac{10}{21}$$
Since, the toy is not replaced there are now 9 even numbered toys and total 20 toys left.
Hence, probability that second toy shows the even number $$=\frac{9}{20}$$
Required probability,
$$\begin{array}{rl}& =\left(\frac{10}{21}\right)\times \left(\frac{9}{20}\right)\\ & =\frac{9}{42}\end{array}$$